Ellipse
Tangents and Common Tangents
Grade 11
Question:
<p>If the tangent at the point \(\left(4\cos\theta, \frac{16\sin\theta}{11}\right)\) to the ellipse \(16x^2 + 11y^2 = 256\) is also tangent to the circle \(x^2 + y^2 - 2x = 15\), then \(\theta\) equals:</p>
<p>(a) \(\frac{\pi}{3}\)</p>
<p>(b) \(\frac{2\pi}{3}\)</p>
<p>(c) \(\frac{5\pi}{6}\)</p>
<p>(d) \(\frac{5\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Find the tangent to the ellipse at the given point, then use the condition that this tangent is also tangent to the circle (distance from center to line equals radius).
<p>The ellipse \(16x^2 + 11y^2 = 256\) can be written as \(\frac{x^2}{16} + \frac{y^2}{\frac{256}{11}} = 1\).</p><p>A point on the ellipse is \(\left(4\cos\theta, \frac{16\sin\theta}{11}\right)\).</p><p>The tangent at this point is: \(\frac{4\cos\theta \cdot x}{16} + \frac{\frac{16\sin\theta}{11} \cdot y}{\frac{256}{11}} = 1\), which simplifies to: \(\frac{x\cos\theta}{4} + \frac{11y\sin\theta}{16} = 1\).</p><p>The circle \(x^2 + y^2 - 2x = 15\) has center \((1, 0)\) and radius \(4\).</p><p>For the line to be tangent to the circle, the distance from the center to the line equals the radius: \(\frac{|\cos\theta/4 - 1|}{\sqrt{\cos^2\theta/16 + 121\sin^2\theta/256}} = 4\).</p><p>Solving this yields \(\theta = \frac{2\pi}{3}\).</p>
Correct Answer: b